I have a change making program I'm writing that asks a user for a monetary amount, then it breaks down that amount into smaller values.
So it uses the Scanner class to elicit a double like 99.99. I want it to use 10.00 as it's largest amount and give 9 x 10.00, 1 x 5.00, 4 x 1.00, 3x .25, etc.
So far I have the following:
Where I'm tripping up is in the division. I'm getting back the wrong amount back (9.998999999999999). I want to divide by 10.00 and get the remainder in a separate variable to then be divided by 5.00, etc.
I don't really get how the whole widening conversion thing works, and I _really_ don't get how to use the % operator in this context to divide by 10.00 and get the remainder of 9.99 back into some type of useful variable that I can then divide?
The book I'm reading explains it, but I seem to be too thick to get it to work on my own.
Thanks for any help.
Joined: Feb 08, 2005
Okay, I think I stumbled onto most of my answer on my own by making the following changes:
I still don't know how to change the return value of "change = (int) cash / 10;" to just a single digit number instead of a decimal value. So if I divide 90.00 with this program it says my change in tens is 9.00, I just want it to say "9". But that's the least of my worries after having figured out the % thing.
Note that you should almost never use floating-point (double or float types) for currency calculations. Floating-point is an approximate method of calculation, for values of widely-varying magnitude. When lots of calculations are done using floating-point, significant errors can build up.
For currency, all values have a fairly similar magnitude, and certainly are significant to the same place. Also, exact answers are required.
Currency calculations should almost always be done using integers (int or long types). A common trick is to use an integer to store the value in the smallest unit of the currency (e.g. cents or pence) and do all calculations in those units. Only convert to the larger unit (e.g. dollars, euros, pounds) for final display to the user.
Betty Rubble? Well, I would go with Betty... but I'd be thinking of Wilma.
I’ve looked at a lot of different solutions, and in my humble opinion Aspose is the way to go. Here’s the link: http://aspose.com
subject: 2 Questions 1) Conversion, 2) Use of the modulo/remainder operator